Answer:
The new force becomes 4 times the initial force.
Explanation:
The force of attraction or repulsion is given by the relation as follows :
[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]
Where
d is the distance between the interacting charges
F is inversely proportional to the distance between charges.
If the distance is halved, d'=(d/2), new force is given by :
[tex]F'=k\dfrac{q_1q_2}{d'^2}\\\\=k\dfrac{q_1q_2}{(\dfrac{d}{2})^2}\\\\=k\dfrac{q_1q_2}{\dfrac{d^2}{4}}\\\\=4\times \dfrac{kq_1q_2}{d^2}\\\\F'=4F[/tex]
So, the new force becomes 4 times the initial force.
